For x(t) = e^(-a|t|), E = integral of e^(-2a|t|) dt is computed as:

2 times integral of e^(-2at) from 0 to infinity

Integral of e^(-2at) from 0 to infinity only

Integral of e^(2at) from -infinity to 0 minus the other half

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2082 Chaitra · 4 marks

Energy or power: x(t) = e^(−a|t|)

Short answer

Compute energy: E = ∫ e^(-2a|t|) dt = 2∫0∞ e^(-2at) dt = 1/a, finite and non-zero. Average power P = lim_{T→∞} (1/(2T))∫_(-T)^T e^(-2a|t|) dt = 0. Therefore x(t)=e^(-a|t|) is an energy signal with E=1/a.

Determine whether the given signal is energy signal or power signal. x(t) = e^(−a|t|), a > 0

$x(t) = e^{-a|t|}$ , $a > 0$ , is a two-sided decaying exponential (even signal).

$|x(t)|^2 = e^{-2a|t|}$ .

Energy: using even symmetry,

\begin{aligned} E &= \int_{-\infty}^{\infty} e^{-2a|t|}\,dt = 2\int_{0}^{\infty} e^{-2at}\,dt \\ &= 2\left[\frac{e^{-2at}}{-2a}\right]_0^{\infty} = 2\cdot\frac{1}{2a} = \frac{1}{a} \end{aligned}

This is finite and non-zero since $a > 0$ .

Power:

P = \lim_{T\to\infty}\frac{1}{2T}\int_{-T}^{T}e^{-2a|t|}\,dt = \lim_{T\to\infty}\frac{1}{2T}\cdot\frac{1-e^{-2aT}}{a} = 0

Answer: $x(t) = e^{-a|t|}$ is an energy signal with $E = 1/a$ and $P = 0$ .

✓ Answer checked by Claude Opus 5.5 (high effort)

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